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Let $U$ be a finite-dimensional $\mathbb{R}$-vector space. I want to show that the following bilinear form over $V:=U\oplus U^\ast$ is symplectic: $$\omega:V\times V\rightarrow\mathbb{R},\quad\omega((u,\varphi),(v,\Psi)):=\varphi(v)-\Psi(u)$$.

A symplectic form is a skew-symmetric bilinear form with trivial kernel. Bilinearity is obvious. From $\varphi(v)-\Psi(u)=-(\Psi(u)-\varphi(v))$, we see that $\omega$ is skew-symmetric.

Now to the kernel. We must show that $\mathrm{ker}(\omega)=\{y\in V: \omega(x,y)=0\}=\{0\}$ for all $x\in V$. So what I did is to see when $\omega_x:=\omega(x,\cdot)=0$ for a given $x\in V$. I find that for $x:=(v,\varphi),y:=(u,\Psi)$ we must have $\varphi(v)=\Psi(u)$ for all $\varphi\in U^\ast,v\in U$, i.e. $\mathrm{ker}(\omega)=\{(u,\varphi)\times(v,\Psi)\in V:\varphi=\Psi\}\neq\{0\}$. However, my lecture notes say that $\omega$ should have a trivial kernel.

Can anybody tell me where I went wrong?

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I think the main issue here might be in the correct phrasing of $\ker \omega$. This link should provide an answer to your question Show skew-symmetric, non-degenerate bilinear form $((a, \varphi),(b, \psi)) \mapsto \langle(a, \varphi),(b, \psi) \rangle := \varphi(b)-\psi(a)$

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